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Official PYQ · 2005Q51Conditional Probability and Bayes TheoremMCQ1 mark

GATE CS 2005 Q51 — Conditional Probability and Bayes Theorem

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Box PP has 22 red balls and 33 blue balls and box QQ has 33 red balls and 11 blue ball. A ball is selected as follows: (i) select a box (ii) choose a ball from the selected box such that each ball in the box is equally likely to be chosen. The probabilities of selecting boxes PP and QQ are 13\dfrac{1}{3} and 23\dfrac{2}{3} respectively. Given that a ball selected in the above process is a red ball, the probability that it came from the box PP is:

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Answer: A

We can solve this using Bayes' theorem. Let PP and QQ denote the events of selecting Box PP and Box QQ respectively, and let RR denote the event of selecting a red ball.

The given probabilities are:

  • P(P)=13P(P) = \dfrac{1}{3} and P(Q)=23P(Q) = \dfrac{2}{3}
  • P(R∣P)=25P(R \mid P) = \dfrac{2}{5} (since Box PP has 22 red and 33 blue balls)
  • P(R∣Q)=34P(R \mid Q) = \dfrac{3}{4} (since Box QQ has 33 red and 11 blue ball)

We need to find the conditional probability P(P∣R)P(P \mid R):

P(P∣R)=P(R∣P)⋅P(P)P(R∣P)⋅P(P)+P(R∣Q)⋅P(Q)P(P \mid R) = \frac{P(R \mid P) \cdot P(P)}{P(R \mid P) \cdot P(P) + P(R \mid Q) \cdot P(Q)}

Substituting the values:

P(P∣R)=25⋅13(25⋅13)+(34⋅23)=215215+12P(P \mid R) = \frac{\dfrac{2}{5} \cdot \dfrac{1}{3}}{\left(\dfrac{2}{5} \cdot \dfrac{1}{3}\right) + \left(\dfrac{3}{4} \cdot \dfrac{2}{3}\right)} = \frac{\dfrac{2}{15}}{\dfrac{2}{15} + \dfrac{1}{2}}P(P∣R)=2154+1530=2151930=215×3019=419P(P \mid R) = \frac{\dfrac{2}{15}}{\dfrac{4 + 15}{30}} = \frac{\dfrac{2}{15}}{\dfrac{19}{30}} = \frac{2}{15} \times \frac{30}{19} = \frac{4}{19}

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